Question

How do you evaluate: `lim_(x rarr 0) xsin(1/x)`?

Answer 1

`lim_(x rarr 0) xsin(1/x) = 0`

Explanation:

Here's an answer using the squeeze or sandwich theorem.

We know that any sine must be between -1 or 1, or in mathematical terms,

`-1 <= sin(theta) <= 1`

If `theta = 1/x` then

`-1 <= sin(1/x) <= 1`

Multiplying both sides by `x`

`-x <= xsin(1/x) <= x` for `x >0`
`-x >= xsin(1/x) >= x` for `x<0`

And since

`lim_(x rarr 0^+) x = lim_(x rarr 0^+) -x = 0`
`lim_(x rarr 0^-) x = lim_(x rarr 0^-) -x = 0`

We can use the squeeze theorem to say

`lim_(x rarr 0) xsin(1/x) = 0`

Answer 2

`lim_(x->0)xsin(1/x) = 0`

Explanation:

Here we are, one year later with a different way to solve this in case somebody comes across this:

`lim_(x->0)xsin(1/x) = lim_(x->0)sin(1/x)/(1/x)`

If we say `u = 1/x` then when `x -> 0`, `u -> oo` so we have

`lim_(u->oo)sin(u)/u`

Remember that the sine is always a number between -1 and 1, so when `u` grows without bound, the denominator becomes much too large and the whole fraction becomes closer and closer to zero. In other words:

`lim_(u->oo)sin(u)/u = 0`